Continuity and Differentiability Test - 16

$$\displaystyle x=\frac{\pi}{4}$$ for $$\displaystyle

x=a\left[\cos{t}+\frac{1}{2}\log{\tan^2{\frac{t}{2}}}\right]$$ and

$$y=a\sin{t}$$

Differentiating w.r.t $$t$$, we get
$$\displaystyle \frac { dx }{ dt } =a\left[ \frac { d }{ dt } \cos { t } +\frac { 1 }{ 2 } \frac { d }{ dt } \left( \log { \tan ^{ 2 }{ \frac { t }{ 2 }  }  }  \right)  \right] $$
$$\displaystyle\frac{dx}{dt}=a\left[-\sin{t}+\frac{1}{\displaystyle\tan{\frac{t}{2}}}\sec^2{\frac{t}{2}}\times\frac{1}{2}\right]$$
$$\displaystyle \frac { dx }{ dt } =a\left[ -\sin { t } +\frac { 1 }{ 2\displaystyle \frac { \sin { { t }/{ 2 } }  }{ \cos { { t }/{ 2 } }  } .\cos ^{ 2 }{ { t }/{ 2 } }  }  \right]$$
$$\displaystyle =a\left[-\sin{t}+\frac{1}{\displaystyle 2\sin{\frac{t}{2}}\cos{\frac{t}{2}}}\right]$$
$$\displaystyle =a\left[-\sin{t}+\frac{1}{\sin{t}}\right]$$
$$=a\left[ \displaystyle \frac { 1-\sin ^{ 2 }{ t }  }{ \sin { t }  }  \right] =a\left[ \displaystyle \frac { \cos ^{ 2 }{ t }  }{ \sin { t }  }  \right] $$
$$\displaystyle\frac{dy}{dt}=a\cos{t}$$
$$\displaystyle\therefore\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{a\cos{t}}{\displaystyle\frac{a\cos^2{t}}{\sin{t}}}=\tan{t}$$
At $$\displaystyle t=\frac{\pi}{4}$$, $$\displaystyle\frac{dy}{dx}=1$$

Taking $$\log$$ on both sides, we get
$$\log{y}={(\tan{x})}^{\displaystyle\tan{x}}\log{\tan{x}}$$
Again taking $$\log$$ on both the sides
$$\log{\log{y}}=[\tan{x}\log{\tan{x}}]+\log{\log{\tan{x}}}$$
Differentiating w.r.t. $$x$$, we get
$$\displaystyle\frac{1}{y\log{y}}\frac{dy}{dx}=\log { \tan { x } \frac { d }{ dx } \left( \tan { x }  \right) +\tan { x } \frac { d }{ dx } \left( \log { \tan { x }  }  \right) +\frac { d }{ dx } \left( \log { \log { \tan { x }  }  }  \right)  } $$

$$\displaystyle \frac{dy}{dx}=\log { \tan { x } .\sec ^{ 2 }{ x } +\tan { x } .\frac { \sec ^{ 2 }{ x }  }{ \tan { x }  } +\frac { \sec ^{ 2 }{ x }  }{ \tan { x } .\log { \tan { x }  }  }  } $$
$$\displaystyle \frac { dy }{ dx } =y\log { y.\sec ^{ 2 }{ x } \left[ \log { \tan { x } +1+\frac { 1 }{ \tan { x } .\log { \tan { x }  }  }  }  \right]  } $$
At $$\displaystyle x=\frac{\pi}{4}$$, $$y=1$$ and $$\log{y}=0$$
So, putting this values in the equation of $$\displaystyle \frac { dy }{ dx } $$, we get
$$\displaystyle\therefore{\left(\frac{dy}{dx}\right)}_{\displaystyle x=\frac{\pi}{4}}=0$$

We have $$x=a\sec^{3}{\theta}$$ and $$y=a\tan^{3}{\theta}$$
Differentiate w. r to $$\theta $$
$$\displaystyle\frac{dx}{d\theta}=3a\sec^{2}{\theta}\displaystyle\frac{d}{d\theta}(\sec{\theta})=3a\sec^{3}{\theta}\tan{\theta}$$
$$\displaystyle\frac{dy}{d\theta}=3a\tan^{2}{\theta}\displaystyle\frac{d}{d\theta}(\tan{\theta})=3a\tan^{2}{\theta}\sec^{2}{\theta}$$
 $$\therefore \displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{d\theta}}{\displaystyle\frac{dx}{d\theta}}=\frac{3a\tan^{2}{\theta}\sec^{2}{\theta}}{3a\sec^{3}{\theta}\tan{\theta}}=\frac{\tan{\theta}}{\sec{\theta}}=\sin{\theta}$$
 $$\displaystyle{\left(\frac{dy}{dx}\right)}_{\displaystyle\theta=\frac{\pi}{3}}=\sin{\frac{\pi}{3}}=\frac{\sqrt{3}}{2}$$

We have $$x=a(\theta-\sin{\theta})$$ and $$y=a(1-\cos{\theta})$$
Differentiate w.r.t $$\theta $$
$$\therefore\displaystyle\frac{dx}{d\theta}=a(1-\cos{\theta})$$ and $$\displaystyle\frac{dy}{d\theta}=a\sin{\theta}$$
or $$\displaystyle\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{d\theta}}{\displaystyle\frac{dx}{d\theta}}$$
$$=\displaystyle\frac{a\sin{\theta}}{a(1-\cos{\theta})}=\frac{2\sin{\left(\displaystyle\frac{\theta}{2}\right)}\cos{\left(\displaystyle\frac{\theta}{2}\right)}}{2\sin^{2}{\left(\displaystyle\frac{\theta}{2}\right)}}=\cot{\left(\displaystyle\frac{\theta}{2}\right)}$$

Given $$\displaystyle y^{x}\:=x^{\sin \:y}$$. 

We have $$\displaystyle f(a)=\log \left [

\frac{a^{2}+ab}{(a+b)a} \right ]=\log

1=0$$ and  $$\displaystyle  f(b)=\log \left [

\frac{b^{2}+ab}{(a+b)b} \right ]=\log1=0$$
$$\Rightarrow f(a)=f(b)=0.$$ Also, it

can be easily seen that $$f(x)$$ is continuous on $$[a,b]$$ and differentiable on

$$[a,b]$$.
Thus all the three conditions of Rolle's

theorem are satisfied. Hence $$\displaystyle

f^{'}(x)=0 $$for at last one value of x in $$[a,b]$$
Now $$\displaystyle f^{'}(x)=0 =

\frac{2x}{x^{2}+ab}-\frac{1}{x}=0\Rightarrow \displaystyle

2x^{2}-(x^{2}+ab)=0=x^{2}=ab$$ or $$\displaystyle

x=\sqrt{ab}$$ which is also known as stationary point.

Let  $$\displaystyle y = x^{\sin^{-1}x}$$ and  $$z = \displaystyle \sin ^{-1}x\Rightarrow \sin z = x$$
we have to find $$\cfrac{dy}{dz}$$
$$\Rightarrow y = (\sin z)^z$$ taking $$\log$$ both side $$\log y = z\log \sin z$$
Differentiating w.r.t $$z$$
$$\displaystyle \frac{1}{y}\frac{dy}{dz}=\log\sin z+z\frac{\cos z}{\sin z}$$
$$\therefore \displaystyle \frac{dy}{dz}=y\left(\log\sin z+z\frac{\cos z}{\sin z}\right)=\displaystyle x^{\sin ^{-1}x}\left [ \log x+\sin ^{-1}x.\frac{\sqrt{\left ( 1-x^{2} \right )}}{x} \right ]$$

Given $$\displaystyle y=(\cot x)^{\sin x}+(\tan  x)^{\cos x}$$

We have $$\displaystyle f(x)=x(x+3)e^{-x/2}$$
$$\displaystyle \therefore f^{'}(x)=(2x+3)e^{-x/2}+(x^{2}+3x)e^{-x/2}\left ( -\frac{1}{2} \right )$$
$$\displaystyle =e^{-x/2}\left [ 2x+3-\frac{1}{2}[x^{2}+3x] \right ]=-\frac{1}{2}[x^{2}-x-6]e^{-x/2}$$
$$f^{'}(x)$$ exist for every value of x in the interval[-3,0]
Hence, $$f(x)$$ is differentiable and hence, continous in the interval $$[-3,0]$$
Also, we have $$\displaystyle f(-3)=f(0)=0\Rightarrow $$All the three conditions of Rolle's theorem are satisfied.
So $$\displaystyle f^{'}(x)=0\Rightarrow  \frac{1}{2}(x^{2}-x-6)e^{-x/6}=0\Rightarrow x^{2}-x-6=0 \Rightarrow  x=3,-2$$
Since, the value $$x=-2$$ lies in the open interval $$[-3,0]$$ the Rolle's theorem is verified.